SIMPLE LOG MATH QUESTION! log (x+6)=1-log(x-5)?

You can apply the log rules easily enough, just convert 1 to log(10).

log(x+6) = log(10) - log(x-5)

log(x+6) = log(10/(x-5))

x+6 = 10/(x-5)

(x+6)(x-5) = 10

x^2 +x -30 = 10

x^2 + x - 40 = 0

Now just shove the equation through the quadratic formula and you'll get your answer.

Hope that helps!

og 5 6+log 5 2 x^2 =log 5 48

log(x+6)=1-log(x-5)

Move all the terms containing a logarithm to the left-hand side of the equation.

log(x+6)+log(x-5)=1

Combine the logarithmic expressions using the product rule of logarithms.

log((x+6)(x-5))=1

Let both sides of the equation be the exponent of base 10.

10^(log((x+6)(x-5)))=10

When the base of a logarithm in the exponent and the base of the exponent are the same, the result is the argument of the logarithm.

(x+6)(x-5)=10

Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.

(x*x+x*-5+6*x+6*-5)=10

Simplify the FOIL expression by multiplying and combining all like terms.

(x^(2)+x-30)=10

Remove the parentheses around the expression x^(2)+x-30.

x^(2)+x-30=10

To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.

x^(2)+x-40=0

Use the quadratic formula to find the solutions. In this case, the values are a=1, b=1, and c=-40.

x=(-b\~(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

Substitute in the values of a=1, b=1, and c=-40.

x=(-1\~((1)^(2)-4(1)(-40)))/(2(1))

Simplify the section inside the radical.

x=(-1\~(161))/(2(1))

Simplify the denominator of the quadratic formula.

x=(-1\~(161))/(2)

First, solve the + portion of \.

x=(-1+~(161))/(2)

Next, solve the - portion of \.

x=(-1-~(161))/(2)

The final answer is the combination of both solutions.

x=(-1+~(161))/(2),(-1-~(161))/(2)_x<Z>APPR<z>5.8443,-6.8443

All answer are incorrect or partly extraordinary! The is a attempt whether the equation holds. because of the fact the quotient x/x is given, the equation is defined for x=0, because of the fact the fact, that there is a real minimize for x goint to 0 of x/x: limit_x->0 x/x=a million. So do away with the quotient a minimum of for x unequal 0. The equation now's x+x²=x². This equation could properly be rearranged by unique arithmetic operations. It supplies x=0. stupid or. What does is recommend? The equation does not carry for x unequal 0. it is it. it could lower back be simplified a million+x=x or a million=0, it is erroneous. For 0 is holds interior the stable variety and interior the minimize attention: 0/0+0=a million+0 unequal 0. The equation does not be valid for any actual numbers!

Hello there,

log (x +6) + log(x -5) = 1

log (x +6)*(x -5) = 1

log(x^2 +x -30) = log 10 (log 10= 1), then cacel the log

x^2 +x -30 =10

x^2 +x -40=0

Δ = b^2 - 4ac = 1^2 - 4(1)(-40)= 161

.................___

x= (-b +/- \/ Δ ) /2a

.................___

x= (-1 +/- \/ 161 ) /2

x1 =(-1 +12.69) /12 = 0.974

x2 =(-1-12.69) / 12 = - 1.14

Therefore, the answers are x1= 0.974; x2 = -1.14

I hope this will help u.

Good luck,

Andrew

log( x+6)+log(x-5) =1

log(x+6)(x-5)=1 (1 is also log10)

log(x^2+x-30)=log 10 (cancel logs)

x^2+x-30=10

x^2+x-40=0

Thats as far as I can go

log(x+6)=log(10)-log(x-5)

log(x+6)=log( 10/(x-5) )

x+6=10/(x-5)

x^2+x-30=10

x^2+x-40=0

x1 = (-1+sqrt(161))/2

x2 = (-1-sqrt(161))/2